解の公式の求め方

■ $a x^{2} + b x + c = 0 (a \neq 0)$ の解の公式の導出

　ただし， $b^{2} - 4 a c ≧ 0$ とする．

$a x^{2} + b x + c = 0$

$a \{x^{2} + \frac{b}{a} x + {(\frac{b}{2 a})}^{2}\} - a {(\frac{b}{2 a})}^{2} + c = 0$

$a \{x^{2} + \frac{b}{a} x + {(\frac{b}{2 a})}^{2}\} = a {(\frac{b}{2 a})}^{2} - c$

$\{x^{2} + \frac{b}{a} x + {(\frac{b}{2 a})}^{2}\} = {(\frac{b}{2 a})}^{2} - \frac{c}{a}$

${(x + \frac{b}{2 a})}^{2} = \frac{b^{2} - 4 a c}{4 a^{2}}$

$x + \frac{b}{2 a} = \pm \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}}$

$x = - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a}$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

　ただし， $b^{2} - a c ≧ 0$ とする．

$a x^{2} + 2 b x + c = 0$

$a \{x^{2} + \frac{2 b}{a} x + {(\frac{b}{a})}^{2}\} - a {(\frac{b}{a})}^{2} + c = 0$

$a \{x^{2} + \frac{2 b}{a} x + {(\frac{b}{a})}^{2}\} = a {(\frac{b}{a})}^{2} - c$

$\{x^{2} + \frac{2 b}{a} x + {(\frac{b}{a})}^{2}\} = {(\frac{b}{a})}^{2} - \frac{c}{a}$

${(x + \frac{b}{a})}^{2} = \frac{b^{2} - a c}{a^{2}}$

$x + \frac{b}{a} = \pm \sqrt{\frac{b^{2} - a c}{a^{2}}}$

$x = - \frac{b}{a} \pm \frac{\sqrt{b^{2} - a c}}{a}$

$x = \frac{- b \pm \sqrt{b^{2} - a c}}{a}$

最終更新日： 2023年7月27日